The average rate of change compares the change in output to the change in input over an interval. From the graph, the two labeled points are approximately (20,58.5) and (26,91.2). The input is days, and the output is widgets produced each day. Use the formula x
2
#x
1
y
2
#y
1
Substituting gives
26#20
91.2#58.5
=
6
32.7
=5.45, which rounds to approximately 5.5. Since this value is an average rate, it means the team's daily production increased by about 5.5 widgets each day during the interval. It is not the total number of additional widgets over the whole interval. Therefore, the correct answer is A.

A logistic function often increases quickly in the middle and then begins to level off as it approaches a maximum carrying capacity. From year 7 to year 10, the graph is still increasing, so the bison population is going up. However, the curve is becoming flatter during this interval. A flatter curve means the slope is getting smaller, so the population is increasing at a slower rate. This behavior is described as "increases slower and slower." It is not decreasing, because the graph continues to move upward. It is also not increasing faster and faster, because the slope is not becoming steeper in this interval. Therefore, the correct interpretation is answer B.

The average rate of change measures how much the output changes per one unit of input. Here, the input is time in hours after 6:00 a.m., and the output is the number of vehicles stopped at the toll booth. The two given points are A=(1,49.4) and B=(4,45.8). Use the average rate of change formula: (y
2
#y
1
)/(x
2
#x
1
). Substituting the coordinates gives (45.8#49.4)/(4#1). The numerator is #3.6, and the denominator is 3, so the average rate of change is #1.2. The negative sign means the number of vehicles decreased over the interval.

The graph shows a curved, downward-opening function. This type of graph is commonly associated with a quadratic polynomial function.
The maximum value of a downward-opening parabola occurs at its highest point, called the vertex.
From the graph, the highest point occurs at approximately:
t=9.5
This means the museum reaches its maximum daily profit approximately:
9.5 " years after opening "
The vertical axis represents daily profit in hundreds of dollars. From the graph, the maximum P(t)-value is approximately:
16.5
Since the profit is measured in hundreds of dollars:
16.5×100=1650
So the maximum daily profit is approximately:
$1,650
Therefore, the correct interpretation is:
" Approximately 9.5 years after opening, a maximum daily profit of approximately " $1,650 " was earned. "

This is an exponential growth problem because the base 1.13 is greater than 1. The function starts with 2,900 bacteria and grows by a factor of 1.13 each hour, which represents a 13% hourly increase. To find the population after 18 hours, substitute t=18 into the function: f(18)=2900(1.13)
18
Evaluating the exponent and multiplying by 2900 gives approximately 26,170.38. Since bacteria are counted as whole organisms, this rounds to about 26,170 bacteria. The other answer choices do not match the exponential model at t=18. Therefore, the correct answer is C.